ユーザー:Kyodaisuu/砂場
自分専用の砂場です。本体にアップする前の下書き等に使います。 Sandbox / 砂場@ヘルプ / 利用者:Kyodaisuu/注目のウィキアン ---- �������� Pair sequence system A program which is supposed to calculate \(f_{\vartheta(\Omega_\omega)+1}(10)\) was posted at googology thread of Japanese BBS 2ch.net (anonymous author).First version of BASIC programModified version of BASIC program It is extention of a program written by the same author. The algorithm is called pair sequence system and the output of this program could be called pair sequence number.Name of the system and the number Pair sequence is a finite sequence of pair of integers, such as (0,0)(1,1)(2,2)(3,3)(3,2). Pair sequence P works as a function from natural numbers to natural numbers, and it can be denoted as Pn, such as (0,0)(1,1)(2,2)(3,3)(3,2)n. The function Pn can be approximated with Hardy hierarchy, and the pair sequence itself represents the ordinal of the approximated Hardy hierarchy, such as \((0,0)(1,1)(2,2)(3,3)(3,2) = \psi(\psi_1(\Omega_2))\). BASIC Code dim A(Infinity):dim B(Infinity):C=9 for D=0 to 9 for E=0 to C A(E)=E:B(E)=E next for F=C to 0 step -1 C=C*C if B(F)=0 then G=1 else G=0 for H=0 to F*G if A(F-H) Calculation up to \(\epsilon_1\) This program uses two arrays A and B which store pair sequence as follows. \= \alpha\ where \(\alpha\) is the corresponding ordinal. In the F loop, first B(F) is checked if B(F)=0 then G=1 else G=0 and when B(F)=0, only the first part is conducted. Substituting G=1, for H=0 to F if A(F-H) This is similar to the program of \(\epsilon_0\) and therefore \begin{eqnarray*} (0,0) &=& 1 \\ (0,0)(1,0) &=& \omega \\ (0,0)(1,0)(2,0) &=& \omega^\omega \\ (0,0)(1,0)(2,0)(3,0) &=& \omega^{\omega^\omega} \\ (0,0)(1,0)(2,0)(3,0)(4,0) &=& \omega^{\omega^{\omega^\omega}} \\ \end{eqnarray*} When B(F)>0, only the latter part is conducted with G=1. Substituting G=1, for K=1 to F if A(F-K) See how it works for (0,0)(1,1) and C=3. F=1 and at K loop, it looks for A(F-K)The author's calculation (154-160, 164-169, 174-184) dim A(Infinity):dim B(Infinity):C=1 for E=0 to 2 A(E)=E:B(E)=E next for F=2 to 0 step -1 if B(F)=0 then G=1 else G=0 for H=0 to F*G if A(F-H) According to the author of the program, at each loop, the sequence of the pairs changes and the corresponding ordinal decreases as follows. It appears to be reasonable, but I am not yet confident with the calculation. I would like to know if googologists here think this calculation is valid. \begin{eqnarray*} (0,0)(1,1)(2,2) &=& \psi(\psi_1(0)) \\ (0,0)(1,1)(2,1) &=& \psi(\Omega) \\ (0,0)(1,1)(2,0)(3,1) &=& \psi(\psi(0)) \\ (0,0)(1,1)(2,0)(3,0) &=& \psi(\omega^\omega) \\ (0,0)(1,1)(2,0)(2,0) &=& \psi(\omega^2) \\ (0,0)(1,1)(2,0)(1,1)(2,0) &=& \psi(\omega \cdot 2) \\ (0,0)(1,1)(2,0)(1,1)(1,1) &=& \psi(\omega+2) \\ (0,0)(1,1)(2,0)(1,1)(1,0)(2,1)(3,0)(2,1) &=& \psi(\omega+1) \cdot \psi(\omega+1) \\ (0,0)(1,1)(2,0)(1,1)(1,0)(2,1)(3,0)(2,0)(3,1)(4,0) &=& \psi(\omega+1) \cdot \psi(\omega) \cdot \psi(\omega) \\ (0,0)(1,1)(2,0)(1,1)(1,0)(2,1)(3,0)(2,0)(3,1)(3,1) &=& \psi(\omega+1) \cdot \psi(\omega) \cdot \psi(1) \\ (0,0)(1,1)(2,0)(1,1)(1,0)(2,1)(3,0)(2,0)(3,1)(3,0)(4,1) &=& \psi(\omega+1) \cdot \psi(\omega) \cdot \psi(0) \cdot \psi(0) \\ (0,0)(1,1)(2,0)(1,1)(1,0)(2,1)(3,0)(2,0)(3,1)(3,0)(4,0) &=& \psi(\omega+1) \cdot \psi(\omega) \cdot \psi(0) \cdot \omega^\omega \\ (0,0)(1,1)(2,0)(1,1)(1,0)(2,1)(3,0)(2,0)(3,1)(3,0)(3,0) &=& \psi(\omega+1) \cdot \psi(\omega) \cdot \psi(0) \cdot \omega^2 \\ (0,0)(1,1)(2,0)(1,1)(1,0)(2,1)(3,0)(2,0)(3,1)(3,0)(2,0)(3,1)(3,0) &=& \psi(\omega+1) \cdot \psi(\omega) \cdot \psi(0) \cdot \omega \cdot 2 \\ (0,0)(1,1)(2,0)(1,1)(1,0)(2,1)(3,0)(2,0)(3,1)(3,0)(2,0)(3,1)(2,0)(3,1) &=& \psi(\omega+1) \cdot \psi(\omega) \cdot (\psi(0) \cdot \omega+\psi(0) \cdot 2) \\ \end{eqnarray*} Calculation with the simplified program which was changed to C=2 was written by the author as follows. The sequence and the corresponding ordinal decrease with each loop of F as follows. \begin{eqnarray*} (0,0)(1,1)(2,2) &=& \psi(\psi_1(0)) \\ (0,0)(1,1)(2,1)(3,1) &=& \psi(\Omega^\Omega) \\ (0,0)(1,1)(2,1)(3,0)(4,1)(5,1)(6,0)(7,1)(8,1) &=& \psi(\omega^{\psi(\omega^{\psi(\omega)})}) \\ (0,0)(1,1)(2,1)(3,0)(4,1)(5,1)(6,0)(7,1)(8,0)(9,1)(10,0)(11,1) &=& \psi(\Omega^{\psi(\Omega^{\psi(\psi(\psi(0)))})}) \\ \text{Denoting } (0,0)(1,1)(2,1)(3,0)(4,1)(5,1)(6,0)(7,1)(8,0)(9,1) &=& X \\ X(10,0)(11,0)(12,0) &=& \psi(\Omega^{\psi(\Omega^{\psi(\psi(\omega^{\omega^\omega}))})}) \\ X(10,0)(11,0)(11,0)(11,0) &=& \psi(\Omega^{\psi(\Omega^{\psi(\psi(\omega^{\omega^3}))})}) \\ X(10,0)(11,0)(11,0)(10,0)(11,0)(11,0)(10,0)(11,0)(11,0) &=& \psi(\Omega^{\psi(\Omega^{\psi(\psi(\omega^{\omega^2 \cdot 3}))})}) \\ X(10,0)(11,0)(11,0)(10,0)(11,0)(11,0)(10,0)(11,0)(10,0)(11,0)(10,0)(11,0) &=& \psi(\Omega^{\psi(\Omega^{\psi(\psi(\omega^{\omega^2 \cdot 3+\omega \cdot 3}))})}) \\ X(10,0)(11,0)(11,0)(10,0)(11,0)(11,0)(10,0)(11,0)(10,0)(11,0)(10,0)(10,0)(10,0) &=& \psi(\Omega^{\psi(\Omega^{\psi(\psi(\omega^{\omega^2 \cdot 3+\omega \cdot 2+3}))})}) \\ \end{eqnarray*} Calculation beyond Bachman-Howard ordinal The author wrote following result. \begin{eqnarray*} (0,0)(1,1)(2,2)(0,0) &=& \psi(\psi_1(0))+1 \\ (0,0)(1,1)(2,2)(1,0) &=& \psi(\psi_1(0)) \omega \\ (0,0)(1,1)(2,2)(2,0) &=& \psi(\psi_1(0) \omega) \\ (0,0)(1,1)(2,2)(3,0) &=& \psi(\psi_1(\omega)) \\ (0,0)(1,1)(2,2)(3,0)(4,0) &=& \psi(\psi_1(\omega^\omega)) \\ (0,0)(1,1)(2,2)(3,0)(4,1) &=& \psi(\psi_1(\psi(0)))=\psi(\psi_1(\epsilon_0)) \\ (0,0)(1,1)(2,2)(3,1) &=& \psi(\psi_1(\Omega)) \\ (0,0)(1,1)(2,2)(3,2) &=& \psi(\psi_1(\Omega_2)) \\ (0,0)(1,1)(2,2)(3,3) &=& \psi(\psi_1(\psi_2(0))) \\ (0,0)(1,1)(2,2)(3,3)(4,4) &=& \psi(\psi_1(\psi_2(\psi_3(0)))) \\ (0,0)(1,1)(2,2)(3,3)...(9,9) &=& \psi(\psi_1(\psi_2(\psi_3(\psi_4(\psi_5(\psi_6(\psi_7(\psi_8(0))))))))) \\ \end{eqnarray*} Therefore, in the loop of "for D=0 to 9" to "next", \(C=f_{\vartheta(\Omega_\omega)}©\) is calculated. Repetition of this loop 10 times results in \(f_{\vartheta(\Omega_\omega)+1}(10)\). Verification code of C Verification of C shows calculation process. It is modified from original algorithm as follows. * D loop eliminated * C is set as constant 2 and does not change * Calculation stops when the length of the sequence exceeds maxlength The result is as follows. * (0,0)(1,1) * (0,0)(1,1)(2,2) * (0,0)(1,1)(2,2)(3,3) * (0,0)(1,1)(2,2)(3,3)(4,4) Sources